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Homework Help: Integrals involving trig functions

  1. Jul 17, 2010 #1
    1. The problem statement, all variables and given/known data

    I need help evaluating the integral Cotx3/10

    I factored out the 1/10 from the integral and am just left with (1/10)*Cotx3

    from here i do not really know what to do. I rewrote it in terms of sine and cosine to get


    I multiply the integral by (1/Sinx3) to get rid of the denominator and am left with


    I factor out a (Cosx2) and am left with


    Rewriting using trig indentities, I get


    rewriting i get


    I solve this integral and get

    (1/10)sinx - (1/10)*((sinx)3/3) + C

    but that is incorrect. The answer is supposed to be


    What went wrong?
  2. jcsd
  3. Jul 17, 2010 #2


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    Well, for one thing, you can't just get rid of parts of the integrand.
    Just to be clear, you are trying to compute

    [tex]\int \cot^3 x\,dx[/tex]

    and not

    [tex]\int \cot x^3\, dx[/tex]


    Hint: Try the substitution u=sin x.
  4. Jul 17, 2010 #3
    Yes, Cot3x
  5. Jul 18, 2010 #4


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    And is it
    [tex]\frac{cot^3 (x)}{10}[/tex]

    Try writing [itex]cot^3(x)[/itex] as

    [tex]\frac{sin^3(x)}{cos^3(x)}= [/tex][tex]\frac{sin^2(x)}{cos^3(x)}cos(x)=[/tex][tex] \frac{1- cos^2(x)}{cos^3(x)} sin(x)[/tex]

    and use the substitution u= cos(x).
  6. Jul 18, 2010 #5
    It is [tex]\frac{cot^3 (x)}{10}[/tex]

    I actually have no idea how to use most of the math tags, so writing it out is hard.

    Cot3x is [tex]\frac{Cos^3(x)}{Sin^3(x)}[/tex]

    I'll try distributing out a cosine factor and converting the rest to sign and see how it goes
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